Given $F^{\mu\nu}$ defined as in QED: $$ F^{\mu\nu} = \begin{bmatrix} 0&E_x&E_y&E_z\\ -E_x&0&-B_z&B_y\\ -E_y&B_z&0&-B_x\\ -E_z&-B_y&B_x&0 \end{bmatrix} $$ when contracting it in the lagrangian as $F^{\mu\nu}F_{\mu\nu}$ how do I compute it? Simply as a product of matrices?


Your first equation involves a common, but misleading, abuse of notation. The left-hand side shows a generic contravariant component $F^{\mu\nu}$ of the field strength tensor and the right hand side shows all components. It would be better to write something like $[F^{\mu\nu}]$ on the left-hand side (simply writing $F$ would be ambiguous as to whether the components are co- or contravariant).

Now, it should be absolutely clear, how to compute $F^{\mu\nu}F_{\mu\nu}$. The Einstein sum convention tells you to sum over all indices that occur once as upper and once as lower index. In other words one has to compute $\sum_\mu \sum_\nu F^{\mu\nu}F_{\mu\nu}$ (here without implicit summation!), where $F^{\mu\nu}$ and $F_{\mu\nu}$ are simply numbers. Don't forget, that $F^{\mu\nu} \ne F_{\mu\nu}$, but you have to contract with the metric tensor first to lower the indices (again using the sum convention): $$ F_{\mu\nu} = F^{\alpha\beta}g_{\mu\alpha}g_{\nu\beta}. $$

  • $\begingroup$ I explicitly chose not give the full answer since the question is close to homework like, and we want to encourage people to work such things out for themselves. Now that it's in the comments it probably won't hurt to edit it in. $\endgroup$ – Sebastian Riese Jul 2 '18 at 21:17
  • $\begingroup$ Fair enough. I have deleted my comment (OP last seen 8 hours ago). $\endgroup$ – Frobenius Jul 2 '18 at 21:25

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